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2 years ago

6

In a second grade class containing 13 girls and 10 boys, 2 students are selected at random to give out the math papers. what is

the probability that the second student chosen is a boy, given that the first one was a girl?

1 answer:

Natali5045456 [20]2 years ago

4 0

The fact that the first chosen student was a girl is already decided, so there is no need to find the probability of that happening. There are a total of 13+10, or 23 students. A girl was already selected, so that leaves 22 students. The chance that a boy is chosen is 10/22, or 5/11.

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Answer:

There were (4x + 2) grapes in the bowl

Step-by-step explanation:

Here, we are interested in calculating the total number of grapes in the bowl.

We have 3 people sharing the total

Damien are x grapes

Jake ate 1 more than twice what Damien and that is (1 + 2x) grapes

Makayla ate 5 fewer grapes than Damien: So what Makayla ate is (x-5) grapes

And now, we have 6 grapes left.

To find the total number of grapes in the bowl, we need to add up all what they ate plus what is left.

Mathematically, that would be;

x + 1 + 2x + x -5 + 6

= 4x + 2

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Step-by-step explanation:

I plugged your data into a graphing calculator, and got a quad reg in the picture.

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Which number is greater than (>) 6.841?

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2 years ago

The percentage of people given an antirheumatoid who suffer severe, moderate, or minor side effects are 11, 20 and 69%, respecti

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Step-by-step explanation:

Since 20 people are administered the medicine and;

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20% of them suffer moderate i.e

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2 years ago

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet

frosja888 [35]

There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

$f(x) = x^5 + \cdots$

Since $x^3$ divides $f(x)$ , and

$f(x) = x^3 p(x)$

where $p(x)$ is degree-2, and we can write it as

$f(x)=x^3 (x^2+ax+b)$

Now, we have

$f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2$

so if $(x - 1)^3$ divides $f(x)^2$ , then $p(x)$ is degree-2, so $p(x)^2$ is degree-4, and we can write

$p(x)^2 = (x-1)^3 q(x)$

where $q(x)$ is degree-1.

Expanding the left side gives

$p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2$

and dividing by $(x-1)^3$ leaves no remainder. If we actually compute the quotient, we wind up with

$\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}$

If the remainder is supposed to be zero, then

$\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}$

Adding these equations together and grouping terms, we get

$(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1$

Then $b=-1-a$ , and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

$2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1$

So, we end up with

$p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}$

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